Physics Friday 3

Kepler's Laws from the Lagrangian Formalism

One can derive Kepler's three laws of planetary motion from Newton's laws; however, they can be derived with perhaps more simplicity using the Lagrangian formalism. In this example, we will be assuming a body of mass m (the “planet”) orbiting a body of much larger mass M (the “star”), with enough larger mass that we can ignore the motion of the larger body (in reality, two bodies orbit their mutual center of gravity; thus the “wobble” of stars due to large planets orbiting them).

First, we will make one use of Newtonian methods. Placing our origin at the star, our planet has position vector and velocity . The force of gravity on the planet is always inward: , (where is the unit vector in the direction of , and r is the magnitude of ). Thus the torque about the origin is as the cross product of parallel or antiparallel vectors is always zero. As torque is zero, the angular momentum is constant. Thus and are both perpendicular to this constant vector , and thus are confined to a plane.

Now, let us use polar coordinates for this plane, with origin at the star.
Thus the kinetic energy of the planet is

We noted before that the force of gravity is . Now, potential energy U is defined as a scalar field such that . Solving this simple differential equation, we find . Thus our Lagrangian is:

First, we consider the Euler-Lagrange equation for θ:


Thus is a constant. Note that, in fact, this is the angular momentum of the planet's orbital motion, so we have
.
Let us consider the line segment connecting the star and the planet. As time progresses, this line segment moves, and will trace out an area. From time to time t, the line will have moved from angle to angle . Using the formula for area enclosed by polar curves:



Thus , and so the line segment connecting the star and planet sweeps out area at a constant rate: this is Kepler's second law.
(Note that Kepler's second law does not depend on U(r) beyond the fact that U is independent of θ; any body in a central force field will obey this law).

Now, let us consider the Euler-Lagrange equation for r:



Or, solving for and substituting,

To solve this equation, we will need several steps. First, let us define . Then , and our equation becomes:
, which is a separable first order equation in v:



We note here that that

, where E is the total energy (which is constant as there are no external forces).
Thus, solving for v,


Here, we are interested in the shape of the orbit, so we change from t to θ. Using
,
we have .
Integrating this,

Defining k=GMm, and using change of variables , we get
. Completing the square in the square root or using a table of integrals, we find that:

Or, solving for μ,

Defining and , and converting back to r, we obtain
.
This is the equation for a conic section of eccentricity ε with a focus at the origin: the orbit is a circle if ε=0, and an ellipse if 0<ε<1).
Thus we have Kepler's first law: planets travel in elliptical orbits with the star at one of the foci.

Now, we just need to establish the third law, which involves the period of closed orbits.

Recalling from our proof of the second law that , we have that the area enclosed in the orbit is
,
where P is the period of the orbit. But the area of an ellipse is given by , where a is the major semi-axis and b the minor semi-axis. From our equation for our ellipse, we see that the major axis runs through the points where and , so


From the definition of the eccentricity of an ellipse, , so

Thus equating our formulas for the area:



or, as it is usually written,

Which is Kepler's third law: the square of the period of the orbit is proportional to the cube of the major semi-axis of the orbit (and independent of the orbiting body's mass).

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